Given the function FX = LG (| X-1 | + | X-5 | - a) 1) when a = 5, find the domain of the function. 2) when the range of the function is r, find the range of A Given the function FX = LG (| X-1 | + | X-5 | - a) 1), when a = 5, find the domain of the function. 2) when the range of the function is r, find the range of A

Given the function FX = LG (| X-1 | + | X-5 | - a) 1) when a = 5, find the domain of the function. 2) when the range of the function is r, find the range of A Given the function FX = LG (| X-1 | + | X-5 | - a) 1), when a = 5, find the domain of the function. 2) when the range of the function is r, find the range of A

1)
When X0
x=5
So x > 11 / 2
So the total is X11 / 2
2)
|x-1|+|x-5|>=|(x-1)-(x-5)|=4
Therefore, in order to make the range R, we must make | X-1 | + | X-5 | - a = 0 solvable,
That is to say, | X-1 | + | X-5 | = a has solution
a>=4

Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

It is proved that: let x = y = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (0) ∵ f (0) ≠ 0, ∵ f (0) = 1, let x = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (y) + F (- y) = 2F (0) · f (y) ∵ f (- y) = f (y), that is, f (x) is even function

It is known that the definition field of function f (x) is R. for any x, y belongs to R, where f (x + y) = f (X-Y) = 2F (x) * f (y), and f (0) is not equal to 0
Proof: 1. F (0) = 1
2. Y = f (x) is even function
3. F (3) = - 2. F (12)

First, f (0) = 1 / 2 instead of 1
It is proved that for any x, y belongs to R, f (x + y) = f (X-Y) = 2F (x) * f (y), and f (0) is not equal to 0
Let x = y = 0
Then f (0 + 0) = f (0-0) = 2F (0) * f (0)
f(0)=1/2
Let x = 0, then f (y) = f (- y), so f (x) is an even function
f(3+3)=2f(3)*f(3)=8
f(6)=8
f(6+6)=2f(6)*f(6)=f(12)=128
I feel that the last question is a paradox, because f (3 + 3) = f (3-3) = f (0) = 1 / 2?

If the odd function f (x) monotonically decreases on (- 1,1), what is the value range of a satisfying the inequality (1-A) + F (1-A Square) less than 0

It should be f (1-A) + F (1-A & sup2;) - 1
It is divided into three parts
1>1-a
a>0
1-a>a²-1
a²+a-2